Limit of x*sin(1/x) as x approaches 0 | Calculus 1 Exercises

We show the limit of xsin(1/x) as x goes to 0 is equal to 0. To do this, we’ll use absolute values and the squeeze theorem, sometimes called the sandwich theorem. We’ll show that |xsin(1/x)| is between 0 and |x|. Then, since 0 and |x| both go to 0 as x goes to 0, we have that |xsin(1/x)| goes to 0 as x approaches 0 by the squeeze theorem. Finally, if the limit of |f(x)| is 0, then the limit of f(x) is 0, and so we can conclude that the limit of xsin(1/x), as x approaches 0, is 0. Review for the AP Calc exam by going through this exam with me! Limit of x*sin(1/x) as x approaches Infinity: Limit of sin(x)/x as x approaches Infinity: Calculus I playlist: ★DONATE★ ◆ Support Wrath of Math on Patreon for early access to new videos and other exclusive benefits: